Class 10 — Electricity

Chapter Overview & Weightage

Electricity is one of the highest-weightage chapters in Class 10 Science. In the CBSE board paper, this chapter alone fetches 6–8 marks across MCQs, short-answer, and long-answer questions. Combined with Magnetic Effects of Current (next chapter), the Physics section is dominated by electricity.

CBSE Class 10 Weightage (Year-by-Year)

Year	Marks from Electricity	Question Types
2024	7	1 MCQ + 1 SA-1 + 1 LA
2023	8	2 MCQ + 1 SA-2 + 1 LA
2022	6	1 SA-1 + 1 LA
2021	7	1 MCQ + 1 SA-2 + 1 LA

The 5-mark long-answer is almost always either a circuit numerical or a derivation (Joule’s law / Ohm’s law).

Key Concepts You Must Know

Ranked by exam frequency:

Ohm’s Law — $V = IR$ . The relation between voltage, current, and resistance. Direct numericals appear every year.

Resistors in Series and Parallel — Series: $R_s = R_1 + R_2 + \cdots$ . Parallel: $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots$ . Compound circuits combining both are the staple of LA questions.

Power and Energy — $P = VI = I^2R = V^2/R$ . Energy consumed: $E = P \times t$ . Unit: kilowatt-hour (1 kWh = $3.6 \times 10^6$ J).

Heating Effect — Joule’s law: $H = I^2Rt$ . Used in heaters, irons, and bulbs.

Resistance and Resistivity — $R = \rho L/A$ . Resistivity depends on material and temperature; resistance depends on dimensions too.

Important Formulas

$V = IR$

V in volts, I in amperes, R in ohms. Use this in every circuit problem.

Series: $R_{\text{eq}} = R_1 + R_2 + R_3$

Parallel: $\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$

For two resistors in parallel: $R_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2}$ .

$P = VI = I^2 R = \frac{V^2}{R}$

Pick the form based on what’s given. If V and R: use $V^2/R$ . If I and R: use $I^2 R$ .

$H = I^2 R t$

Heat generated in a resistor (in joules) when current $I$ flows for time $t$ .

Solved Previous Year Questions

PYQ 1 — CBSE 2024, 5 Marks

Two resistors of $5\,\Omega$ and $10\,\Omega$ are connected in parallel across a $12$ V battery. Find (a) total resistance, (b) total current, (c) current through each resistor, and (d) power dissipated by each.

(a) $R_{\text{eq}} = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} \approx 3.33\,\Omega$ .

(b) Total current: $I = V/R_{\text{eq}} = 12/3.33 \approx 3.6$ A.

(c) Voltage across each is 12 V (parallel). Current in $5\,\Omega$ : $I_1 = 12/5 = 2.4$ A. Current in $10\,\Omega$ : $I_2 = 12/10 = 1.2$ A. Sum = 3.6 A. ✓

(d) Power in $5\,\Omega$ : $P_1 = V^2/R_1 = 144/5 = 28.8$ W. Power in $10\,\Omega$ : $P_2 = 144/10 = 14.4$ W.

PYQ 2 — CBSE 2023, 3 Marks

State Joule’s law of heating and use it to derive the SI unit of electrical energy.

Joule’s law: heat produced in a conductor is directly proportional to (i) square of current, (ii) resistance, (iii) time. Mathematically, $H = I^2 R t$ .

Energy = Power × time = $V \times I \times t =$ joule (J). For larger units: 1 kWh = 1000 W × 3600 s = $3.6 \times 10^6$ J.

PYQ 3 — CBSE 2022, 5 Marks

Find the equivalent resistance and current drawn from the source for a circuit where two parallel branches (each branch has a $4\,\Omega$ + $6\,\Omega$ in series) are connected to a $20$ V source.

Each branch: $R_{\text{branch}} = 4 + 6 = 10\,\Omega$ . Two such branches in parallel:

$R_{\text{eq}} = 10/2 = 5\,\Omega$ . Total current: $I = 20/5 = 4$ A.

Difficulty Distribution

Difficulty	% of Marks	Sub-topics
Easy	25%	Ohm’s law direct, simple series/parallel
Medium	50%	Compound circuits, power calculations, Joule’s law applications
Hard	25%	Multi-step circuits, derivations, household wiring

Expert Strategy

Week 1 — Ohm’s law and basic circuits. Do every NCERT exercise problem with a clean diagram. Label every node before computing.

Week 2 — Power, energy, kWh problems. Practice converting between watts, kilowatt-hours, and joules. CBSE loves “monthly electricity bill” word problems.

Week 3 — Mixed circuits + Joule’s heating. Tackle compound circuits where you must reduce step-by-step. Memorise the parallel-of-two formula $R_1 R_2/(R_1 + R_2)$ — saves 30 seconds every time.

Topper’s tip: Always draw the circuit diagram clearly and label currents and voltages before computing. Most lost marks come from messy diagrams, not wrong formulas. CBSE awards 1 mark just for a clean labelled diagram in 5-mark questions.

Common Traps

Trap 1: Confusing series and parallel rules.

Series: same current, voltages add. Parallel: same voltage, currents add. Mixing these up gives the wrong equivalent resistance every time.

Trap 2: Assuming bulbs in series glow equally bright.

Power = $I^2 R$ , and current is the same in series, so bulbs with higher resistance glow brighter in series. In parallel, voltage is shared equally, so $P = V^2/R$ — lower resistance glows brighter.

Trap 3: Forgetting to convert kWh to joules in heat problems.

CBSE numericals often mix units: voltage in V, current in A, time in hours, energy asked in joules. Always convert to SI: time in seconds, energy in joules.

Trap 4: Using P = VI when only resistance is given.

If the problem gives only $V$ and $R$ , use $P = V^2/R$ . Calculating $I$ first wastes time and adds error.

Trap 5: Adding parallel resistances directly.

$R_p \neq R_1 + R_2$ for parallel. Always invert, add reciprocals, invert back. Or use the two-resistor shortcut $R_1 R_2 / (R_1 + R_2)$ .